Answer:
Potential energy stored in the spring is given as
U = 0.625 J
Explanation:
As we know that the potential energy stored in the spring is given as
[tex]U = \frac{1}{2}kx^2[/tex]
so we have
K = 20 N/m
also the spring is compressed by
x = 0.25 m
so we have
[tex]U = \frac{1}{2}(20)(0.25)^2[/tex]
[tex]U = 0.625 J[/tex]
